Answer:
(b) BeF2 > OF2 > CH3OH
Explanation:
The degree and type of intermolecular forces present in a substance influences its vapour pressure considerably. The greater the magnitude and strength of intermolecular forces in the substance, the lower the vapour pressure of the substance.
BeF2 molecules are held together by weak vanderwaals forces hence BeF2 will exhibit the least degree of intermolecular interaction and have the highest vapour pressure. OF2 molecules are bound together by dipole interactions hence it will exhibit a lower vapour pressure compared to BeF2. CH3OH molecules form hydrogen bonds with water molecules hence it will exhibit the least vapour pressure among the trio.
Bromine attracts electrons more strongly. Cesium is In fact the least electro negative element.
Sodium is more likely to lose an electron because is is less electro negative. Strong electronegativity make the element want more electrons. Sodium has loose electrons with a lower electronegativity so it gives it up easier.
Answer:
Current randomly flows around the circuitt
Answer:
Explanation:
- The formulae used in solving the problem ; ΔTf = i × Kf × m
- where ΔTf = depression in freezing point ,
- i = vant hoff factor 1 for non dissociable solutes ,
- Kf = freezing point constant for solvent
Appropriate substitution was done as shown in the attached file.
0.0367 seconds = 36.7 milliseconds
